Answer:
0.657 seconds
Explanation:
speed of wave= wavelength / time period
so
time period= wavelength / speed
= 4.6/7
=0.657 sec
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
<u>λ = 5.2 x 10⁻⁷ m = 520 nm</u>
Answer:
The loop penetrate 4 cm into the magnetic field.
Explanation:
Given that,
Width w= 5 cm
Length L= 10 cm
mass m = 40 g
Resistance R = 20 mΩ
Initial velocity = 1 m/s
Magnetic field = 2 T
We need to calculate the induced emf
Using formula of emf

Put the value into the formula


We need to calculate the current
Using Lenz's formula



We need to calculate the force
Using formula of force


Put the value into the formula


We need to calculate the acceleration
Using formula of acceleration

Put the value in to the formula


We need to calculate the distance
Using equation of motion





Hence, The loop penetrate 4 cm into the magnetic field.
Scientists think the solar system formed around 4.6 billion years ago from a cloud of dust and gas known as a solar nebula. Gravity collapsed the material on itself as it began to spin. This formed the sun in the middle. The planets formed when rock joined together in clumps and became bigger and bigger. And thus, our solar system was formed. Hope this helped
Answer:
B. 1700 Hz, 5100 Hz
Explanation:
Parameters given:
Length of ear canal = 5.2cm = 0.052 m
Speed of sound in warm air = 350 m/s
The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:
f(m) = m * (v/4L)
Where m is an odd integer;
v = velocity
L = length of the tube
Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).
f(1) = 1 * [350/(4*0.052)]
f(1) = 1682.69 Hz
Approximately, f(1) = 1700 Hz
f(3) = 3 * [350/(4*0.052)]
f(3) = 5048 Hz
Approximately, f(3) = 5100 Hz